Further results on the group inverse of some anti-triangular block matrices

Abstract

Suppose ℜ is a right Ore domain with unity 1. In this paper, we investigate the existence of the group inverse of some anti-triangular block matrices over ℜ and obtain the sufficient and necessary conditions for such existence. Further, the representations of the group inverse for the following two classes are given. (i) , where CA=C; (ii) , where B ^♯ exists and BA=BAB ^♯ B.

The results extend the earlier works of Liu et al. (in Appl. Math. Comput. 218:8978–8986, 2012) and Zhao et al. (in E. J. Linear Algebra 21:63–75, 2010). Some results in special cases are also generalized to any ring.

1 Introduction

For a square matrix A, the matrix X is called the group inverse of A if X satisfies the matrix equations

$$AXA=A, XAX=X\quad \mbox{and}\quad AX=XA. $$

It is well known that if the group inverse X exists, it is unique, and is denoted by A ^♯. Let A ^π be I−AA ^♯.

The group inverse of block matrix is a very useful tool in many fields, such as iterative methods, Markov chains, singular differential and difference equations, see [1–8].

In 1983, in the context of differential equations, Campbell et al. [9] proposed the problem of finding the representation for the Drazin inverse (group inverse) of the anti-triangular block matrix . This problem remains open. However, there are many results in some special cases, see [10, 13–18, 20–22]. It is important to study them in a larger ring, see for example [11, 12, 19, 26].

Liu et al. [10] studied this problem under the conditions A ²=A,CA=C over complex fields. In this paper, we not only delete the condition A ²=A but also solve it for the matrices over right Ore domains by using matrix equations. This also generalizes the results of Ge et al. [27].

On the other hand, in [20], Zhao et al. characterized the existence and the representation of group inverse for block matrix over skew fields under the condition that B ^♯ exists and BAB ^π=0. In this paper, we extend these results to right Ore domains. Some results in special cases are studied over rings with unity 1.

In this paper, let R be a ring with unity 1. A ring is called a right Ore domain (denoted by ℜ) if it possesses no zero divisors and every two elements of the ring have a right common multiple. Integral rings, polynomial rings in an indeterminate over field, noncommutative principal ideal domains and so on are right Ore domains. A left Ore domain is defined similarly. Every right (left) Ore domain ℜ can be embedded in the skew field (denoted by K _ℜ) of quotients of itself. More details are found in [23–25]. Let ℜ^m×n (respectively, R ^m×n) be the set of all m×n matrices over ℜ (respectively, R). The rank of a matrix A∈ℜ^m×n (denoted by r(A)) is defined as the rank of A over K _ℜ, i.e., the maximum order of all invertible subblocks of A over K _ℜ. For convenience, we suppose the right Ore domain ℜ has identity 1.

2 Some lemmas

The following three lemmas will be used in the paper.

Lemma 1

[12]

Let A∈ℜ^n×n. Then the following are equivalent:

• (i) A ^♯ exists;

• (ii) A ² X=A for some X∈ℜ^n×n. In this case, A ^♯=AX ²;

• (iii) YA ²=A for some Y∈ℜ^n×n. In this case, A ^♯=Y ² A.

Lemma 2

[11, 26]

Let A∈R ^n×n. Then the following are equivalent:

• (i) A ^♯ exists;

• (ii) A ² X=A,YA ²=A for some X,Y over R. In this case, A ^♯=Y ² A=AX ²=YAX.

Lemma 3

Let A,B∈R ^n×n. If BAB ^π=0, B ^♯ and (AB ^π)^♯ exist, then

• (i) B ^♯ AB ^π=0, (AB ^π)^♯ B=0, B(AB ^π)^♯=0;

• (ii) A(AB ^π)^♯=(AB ^π)^♯ AB ^π;

• (iii) A(AB ^π)^♯ AB ^π=AB ^π.

Proof

(i) B ^♯ AB ^π=(B ^♯)² BAB ^π=0, (AB ^π)^♯ B=((AB ^π)^♯)² AB ^π B=0, similarly, B(AB ^π)^♯=0.

(ii) A(AB ^π)^♯=AB ^π(AB ^π)^♯=(AB ^π)^♯ AB ^π.

(iii) From (ii), we can easily show that (iii) holds. □

3 Main results

The following is the main result in this note.

Theorem 1

Let , where A∈ℜ^n×n,B∈ℜ^n×m,C∈ℜ^m×n. If CA=C, then

• (i) M ^♯ exists if and only if (CB)^♯ and A ^♯ exist, A ^π B(CB)^π=0;

• (ii) If M ^♯ exists, then , where

  $$\begin{aligned} M_{1} =&A^{\sharp}\,{-}\,(A^{\sharp})^{2}B(CB)^{\pi}C\,{-}\,A^{\sharp}B(CB)^{\sharp }C-A^{\sharp}B(CB)^{\pi}C\,{-}\,A^{\pi}B[(CB)^{\sharp}]^{2}C, \\ M_{2} =&(A^{\sharp})^{2}B(CB)^{\pi}+A^{\pi}B[(CB)^{\sharp}]^{2}+A^{\sharp }B(CB)^{\sharp}+A^{\pi}B(CB)^{\sharp}, \\ M_{3} =&(CB)^{\pi}C+(CB)^{\sharp}C, \\ M_{4} =& -(CB)^{\sharp}. \end{aligned}$$

Proof

(i) The “only if” part.

Since CA ²=CA=C, we have

$$ \left ( \begin{array}{c@{\quad}c}I&B\\ 0&I \end{array} \right )\left ( \begin{array}{c@{\quad}c}I&0\\ C&I \end{array} \right ) \left ( \begin{array}{c@{\quad}c}A^{2}&AB-BCB\\ 0&CBCB \end{array} \right )= \left ( \begin{array}{c@{\quad}c}A^{2}+BC&AB\\ C&CB \end{array} \right )= M^{2} $$

(1)

By Lemma 1, M ^♯ exists if and only if M=YM ² for some Y∈ℜ^(n+m)×(n+m).

Let

$$Y\left ( \begin{array}{c@{\quad}c}I&B\\ 0&I \end{array} \right )\left ( \begin{array}{c@{\quad}c}I&0\\ C&I \end{array} \right ) =\left ( \begin{array}{c@{\quad}c}Y_{1}&Y_{2}\\ Y_{3}&Y_{4} \end{array} \right ), $$

where Y ₁∈ℜ^n×n, so

$$\left ( \begin{array}{c@{\quad}c}Y_{1}&Y_{2}\\ Y_{3}&Y_{4} \end{array} \right ) \left ( \begin{array}{c@{\quad}c}A^{2}&AB-BCB\\ 0&CBCB \end{array} \right ) =\left ( \begin{array}{c@{\quad}c}A&B\\ C&0 \end{array} \right ), $$

it follows that Y ₁ A ²=A, by Lemma 1, so A ^♯ exists.

According to Lemma 1, M ^♯ exists if and only if there exists a matrix such that M ² X=M, where X ₁∈ℜ^n×n.

By (1), we have

$$\left ( \begin{array}{c@{\quad}c}I&B\\ 0&I \end{array} \right )\left ( \begin{array}{c@{\quad}c}I&0\\ C&I \end{array} \right ) \left ( \begin{array}{c@{\quad}c}A^{2}&AB-BCB\\ 0&CBCB \end{array} \right )X=M. $$

Hence,

$$\left ( \begin{array}{c@{\quad}c}A^{2}&AB-BCB\\ 0&CBCB \end{array} \right )\left ( \begin{array}{c@{\quad}c}X_{1}&X_{2}\\ X_{3}&X_{4} \end{array} \right )=\left ( \begin{array}{c@{\quad}c}I&0\\ -C&I \end{array} \right ) \left ( \begin{array}{c@{\quad}c}I&-B\\ 0&I \end{array} \right )M. $$

Namely,

$$ {A^{2}X_{1}+(AB-BCB)X_{3}=A-BC}, $$

(2)

$$ {A^{2}X_{2}+(AB-BCB)X_{4}=B}, $$

(3)

$$ {CBCBX_{3}=CBC}, $$

(4)

$$ {CBCBX_{4}=-CB}. $$

(5)

From (5) and Lemma 1, we know (CB)^♯ exists. By (3), we have −A ^π BCBX ₄=A ^π B. i.e.,

$$-A^{\pi}B(CB)^{\sharp}CBCBX_{4}=A^{\pi}B. $$

Substitute (5) into the above equation, we have A ^π B(CB)^♯ CB=A ^π B. Therefore, A ^π B(CB)^π=0.

The “if” part.

Let $X_{1}=A^{\sharp}-A^{\sharp^{2}}B(CB)^{\pi}C-A^{\sharp}B(CB)^{\sharp}C$, $X_{2}=A^{\sharp^{2}}B(CB)^{\pi}+A^{\sharp}B(CB)^{\sharp}$, X ₃=(CB)^♯ C,X ₄=−(CB)^♯.

Note that A ^π B(CB)^π=0. It is easy to verify that (2)–(5) hold. This implies M=M ² X has a solution, so M ^♯ exists.

(ii) By Lemma 1, A ^π B(CB)^π=0 and CA ^♯=CAA ^♯=CA ² A ^♯=CA=C, the expression of M ^♯ can be obtained from M ^♯=MX ². Next we can compute that

$$\begin{aligned} M^{\sharp} =& \left ( \begin{array}{c@{\quad}c}A&B\\ C&0 \end{array} \right ) \left ( \begin{array}{c@{\quad}c}X_{1}&X_{2}\\ X_{3}&X_{4} \end{array} \right ) \left ( \begin{array}{c@{\quad}c}X_{1}&X_{2}\\ X_{3}&X_{4} \end{array} \right ) \\ =& \left ( \begin{array}{c@{\quad}c}A&B\\ C&0 \end{array} \right ) \\ &{} \times \left ( \begin{array}{c@{\quad}c}A^{\sharp}-A^{\sharp^{2}}B(CB)^{\pi}C-A^{\sharp }B(CB)^{\sharp}C& A^{\sharp^{2}}B(CB)^{\pi}+A^{\sharp}B(CB)^{\sharp}\\ (CB)^{\sharp }C&-(CB)^{\sharp} \end{array} \right ) \\ &{} \times \left ( \begin{array}{c@{\quad}c}X_{1}&X_{2}\\ X_{3}&X_{4} \end{array} \right ) \\ =& \left ( \begin{array}{c@{\quad}c}AA^{\sharp}-A^{\sharp}B(CB)^{\pi}C+A^{\pi}B(CB)^{\sharp}C& A^{\sharp}B(CB)^{\pi}-A^{\pi}B(CB)^{\sharp}\\ (CB)^{\pi}C&CB(CB)^{\sharp} \end{array} \right ) \\ &{} \times \left ( \begin{array}{c@{\quad}c}X_{1}&X_{2}\\ X_{3}&X_{4} \end{array} \right ) \\ =& \left ( \begin{array}{c@{\quad}c}A^{\sharp}-A^{\sharp^{2}}B(CB)^{\pi}C-A^{\sharp }B(CB)^{\sharp}C& A^{\sharp^{2}}B(CB)^{\pi}+A^{\pi}B((CB)^{\sharp})^{2}\\ -A^{\sharp}B(CB)^{\pi}C-A^{\pi}B((CB)^{\sharp})^{2}C&+A^{\sharp }B(CB)^{\sharp}+A^{\pi}B(CB)^{\sharp}\\ (CB)^{\pi}C+(CB)^{\sharp}C&-(CB)^{\sharp} \end{array} \right ) \\ =& \left ( \begin{array}{c@{\quad}c}M_{1}&M_{2}\\ M_{3}&M_{4} \end{array} \right ). \end{aligned}$$

 □

Example for Theorem 1: Let ℜ be the integer ring, and let , where

$$A=\left ( \begin{array}{c@{\quad}c@{\quad}c}1&0&0\\ 0&-1&0\\ 0&0&0 \end{array} \right ),\qquad B=\left ( \begin{array}{c@{\quad}c} 1&0\\ 1&3\\ 2&0 \end{array} \right ),\qquad C=\left ( \begin{array}{c@{\quad}c@{\quad}c}1&0&0\\ 2&0&0 \end{array} \right ). $$

It is easy to verify A ²≠A,CA=C. Furthermore, A ^♯ and (CB)^♯ exist.

By computation,

$$\begin{aligned} & A^{\sharp}=\left ( \begin{array}{c@{\quad}c@{\quad}c} 1&0&0\\ 0&-1&0\\ 0&0&0 \end{array} \right ),\qquad A^{\pi}=\left ( \begin{array}{c@{\quad}c@{\quad}c}0&0&0\\ 0&0&0\\ 0&0&1 \end{array} \right ),\qquad (CB)^{\sharp}=CB=\left ( \begin{array}{c@{\quad}c}1&0\\ 2&0 \end{array} \right ),\\ & (CB)^{\pi}=\left ( \begin{array}{c@{\quad}c}0&0\\ -2&1 \end{array} \right ), \end{aligned}$$

so A ^π B(CB)^π=0. By Theorem 1, M ^♯ exists and

$$M^{\sharp}=\left ( \begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 0&0&0&1&0\\ 7&-1&0&-13&3\\ -2&0&0&4&0\\ 1&0&0&-1&0\\ 2&0&0&-2&0 \end{array} \right ). $$

Similarly, we can prove the counterpart of Theorem 1.

Theorem 2

Let , where A∈ℜ^n×n,B∈ℜ^n×m,C∈ℜ^m×n. If AB=B, then

• (i) M ^♯ exists if and only if (CB)^♯ and A ^♯ exist and (CB)^π CA ^π=0;

• (ii) If M ^♯ exists, then , where

  $$\begin{aligned} M_{1} =&A^{\sharp}\,{-}\,B(CB)^{\pi}CA^{\sharp}\,{-}\,B(CB)^{\pi}C(A^{\sharp })^{2}\,{-}\,B(CB)^{\sharp}CA^{\sharp}\,{-}\,B[(CB)^{\sharp}]^{2}CA^{\pi}, \\ M_{2} =&B(CB)^{\sharp}+B(CB)^{\pi}, \\ M_{3} =&(CB)^{\sharp}CA^{\sharp}+(CB)^{\sharp}CA^{\pi}+[(CB)^{\sharp }]^{2}CA^{\pi}+(CB)^{\pi}C(A^{\sharp})^{2}, \\ M_{4} =&-(CB)^{\sharp}. \end{aligned}$$

Next we consider a special case of Theorems 1 and 2, and investigate it over any ring.

Theorem 3

Let , where A∈R ^n×n,B∈R ^n×m,C∈R ^m×n. If AB=B,CA=C, then

• (i) M ^♯ exists if and only if (CB)^♯ and A ^♯ exist;

• (ii) If M ^♯ exists, then

  $$M^{\sharp}=\left ( \begin{array}{c@{\quad}c} A^{\sharp}-2B(CB)^{\pi}C-B(CB)^{\sharp}C&B(CB)^{\sharp }+B(CB)^{\pi}\\ (CB)^{\sharp}C+(CB)^{\pi}C&-(CB)^{\sharp} \end{array} \right ). $$

Proof

(i) The “only if” part.

If M ^♯ exists, then by Lemma 2 there exist matrices X and Y over R such that M=M ² X and M=YM ².

By AB=B and CA=C, we have

$$\begin{aligned} \left ( \begin{array}{c@{\quad}c}I&0\\ C&I \end{array} \right ) \left ( \begin{array}{c@{\quad}c}A^{2}&0\\ -CBC&CBCB \end{array} \right ) \left ( \begin{array}{c@{\quad}c}I&B\\ 0&I \end{array} \right )\left ( \begin{array}{c@{\quad}c}I&0\\ C&I \end{array} \right ) =\left ( \begin{array}{c@{\quad}c}A^{2}+BC&B\\ C&CB \end{array} \right ) =M^{2}. \end{aligned}$$

Let

$$X=\left ( \begin{array}{c@{\quad}c}I&0\\ -C&I \end{array} \right )\left ( \begin{array}{c@{\quad}c}I&-B\\ 0&I \end{array} \right ) \left ( \begin{array}{c@{\quad}c}X_{1}&X_{2}\\ X_{3}&X_{4} \end{array} \right )\left ( \begin{array}{c@{\quad}c}I&B\\ 0&I \end{array} \right ). $$

Then

$$\left ( \begin{array}{c@{\quad}c}A^{2}&0\\ -CBC&CBCB \end{array} \right )\left ( \begin{array}{c@{\quad}c}X_{1}&X_{2}\\ X_{3}&X_{4} \end{array} \right ) =\left ( \begin{array}{c@{\quad}c}A&0\\ 0&-CB \end{array} \right ). $$

From above, we get

$$ {A^{2}X_{1}=A}, $$

(6)

$$ {A^{2}X_{2}=0}, $$

(7)

$$ {-CBCX_{1}+CBCBX_{3}=0}, $$

(8)

$$ {-CBCX_{2}+CBCBX_{4}=-CB}. $$

(9)

It is easy to get

$$\left ( \begin{array}{c@{\quad}c}I&B\\ 0&I \end{array} \right )\left ( \begin{array}{c@{\quad}c}I&0\\ C&I \end{array} \right ) \left ( \begin{array}{c@{\quad}c}A^{2}&B-BCB\\ 0&CBCB \end{array} \right )=M^{2}. $$

From above equations and M=YM ², let , we have

$$ {Y_{1}A^{2}=A} , $$

(10)

$$ {Y_{1}(B-BCB)+Y_{2}CBCB=B}, $$

(11)

$$ {Y_{3}A^{2}=C}, $$

(12)

$$ {Y_{3}(B-BCB)+Y_{4}CBCB=0}. $$

(13)

By Lemma 2, (6) and (10) imply A ^♯ exists.

By (7) and (12), we have CX ₂=0 and Y ₃ B=CB. Substitute these identities into (9) and (13) respectively, we get CBCBX ₄=−CB and CB=(I−Y ₄)CBCB.

By Lemma 1, (CB)^♯ exists.

The ‘if’ part. Let

$$\begin{aligned} &X_{1}=A^{\sharp}, X_{2}=0, X_{3}=(CB)^{\sharp}C, X_{4}=-(CB)^{\sharp}. \\ &Y_{1}=A^{\sharp}, Y_{2}=B(CB)^{\sharp}, Y_{3}=C, Y_{4}=CB(CB)^{\sharp}-(CB)^{\sharp}. \end{aligned}$$

It is easy to verify (6)–(13) hold. That implies M=M ² X and M=YM ² have solutions.

From Lemma 2, we know M ^♯ exists.

(ii) By Lemma 2, the expression of M ^♯ can get from M ^♯=YMX. □

Example for Theorem 3: Let Z be the integer ring, and let be a matrix over Z/(6Z), where

$$A=\left ( \begin{array}{c@{\quad}c@{\quad}c}1&-2&0\\ 0&1&0\\ 0&0&0 \end{array} \right ),\qquad B=\left ( \begin{array}{c@{\quad}c}3&1\\ 3&3\\ 0&0 \end{array} \right ),\qquad C=\left ( \begin{array}{c@{\quad}c@{\quad}c} 3&1&0\\ 3&2&0 \end{array} \right ). $$

It is easy to verify that A ²≠A, AB=B, CA=C. Furthermore, A ^♯ and (CB)^♯ exist.

By computation,

$$A^{\sharp}=\left ( \begin{array}{c@{\quad}c@{\quad}c}1&2&0\\ 0&1&0\\ 0&0&0 \end{array} \right ),\qquad (CB)^{\sharp}=CB=\left ( \begin{array}{c@{\quad}c}0&0\\ 3&3 \end{array} \right ). $$

By Theorem 3, M ^♯ exists and

$$M^{\sharp}=\left ( \begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 1&1&0&3&1\\ 0&-2&0&3&3\\ 0&0&0&0&0\\ 3&1&0&0&0\\ 3&2&0&3&3 \end{array} \right ). $$

Remark 1

From Theorem 1 and 2, we can obtain Theorem 3.1 and 3.2 of [27] and Theorem 2.1 and 2.2 of [10]. From Theorem 3, we can also obtain the Corollary 2.2 of [10]. In above two examples, we especially point that A ²≠A. This shows that the generalizations are true.

The following results extend the corresponding works of Zhao et al. [20].

Theorem 4

Let , where A,B∈ℜ^n×n. If B ^♯ exists and BAB ^π=0. Then

• (i) M ^♯ exists if and only if (AB ^π)^♯ exists.

• (ii) If M ^♯ exists, then , where

  $$\begin{aligned} M_{1} =&B^{\pi}A(B^{\sharp})^{2}-(AB^{\pi})^{\sharp}AB^{\pi}A(B^{\sharp })^{2}+(AB^{\pi})^{\sharp}; \\ M_{2} =&-B^{\pi}A(B^{\sharp})^{2}AB^{\sharp}+(AB^{\pi})^{\sharp}AB^{\pi }A(B^{\sharp})^{2}AB^{\sharp} -(AB^{\pi})^{\sharp}AB^{\sharp}+B^{\sharp}; \\ M_{3} =&B^{\sharp}; \\ M_{4} =&-B^{\sharp}AB^{\sharp}. \end{aligned}$$

Proof

The “only if” part of (i).

It is easy to get

$$M=\left ( \begin{array}{c@{\quad}c}A&B\\ B&0 \end{array} \right ) =\left ( \begin{array}{c@{\quad}c}B^{\pi}A&B\\ B&0 \end{array} \right ) \left ( \begin{array}{c@{\quad}c}I&O\\ B^{\sharp}A&I \end{array} \right ), $$

$$M^{2}=\left ( \begin{array}{c@{\quad}c}A^{2}+B^{2}&AB\\ BA&B^{2} \end{array} \right ) =\left ( \begin{array}{c@{\quad}c}AB^{\pi}A+B^{2}&AB\\ 0&B^{2} \end{array} \right ) \left ( \begin{array}{c@{\quad}c}I&O\\ B^{\sharp}A&I \end{array} \right ). $$

Since M ^♯ exists, from Lemma 1 we know YM ²=M has a solution.

Let .

We have

$$ {Y_{1}AB^{\pi}A+Y_{1}B^{2}=B^{\pi}A}, $$

(14)

$$ {Y_{1}AB+Y_{2}B^{2}=B}, $$

(15)

$$ {Y_{3}AB^{\pi}A+Y_{3}B^{2}=B}, $$

(16)

$$ {Y_{3}AB+Y_{4}B^{2}=0}. $$

(17)

By (14), we have Y ₁(AB ^π)²=B ^π AB ^π=AB ^π. From Lemma 1, we know (AB ^π)^♯ exists. Next, we prove the sufficiency of (i) and the expression of (ii):

Let . By Lemma 3, the sufficiency of (i) and the expression of M ^♯ are similar to the proof in [20].

$$MX=\left ( \begin{array}{c@{\quad}c}A(AB^{\pi})^{\sharp}+BB^{\sharp}&-A(AB^{\pi})^{\sharp }AB^{\sharp}+B^{\pi}AB^{\sharp} \\ 0&BB^{\sharp} \end{array} \right )=XM. $$

It is easy to verify that XMX=M,MXM=M.

So X=M ^♯. □

Similarly, we state the symmetrical result of Theorem 4.

Theorem 5

Let , where A,B∈ℜ^n×n. If B ^♯ exists and B ^π AB=0, then

• (i) M ^♯ exists if only if (B ^π A)^♯ exists.

• (ii) If M ^♯ exists, then , where

  $$\begin{aligned} M_{1} =&(B^{\sharp})^{2}AB^{\pi}-(B^{\sharp})^{2}AB^{\pi}A(B^{\pi }A)^{\sharp}+(B^{\pi}A)^{\sharp}; \\ M_{2} =&B^{\sharp}; \\ M_{3} =&-B^{\sharp}A(B^{\sharp})^{2}AB^{\pi}+B^{\sharp}A(B^{\sharp })^{2}AB^{\pi}A(B^{\pi}A)^{\sharp} -B^{\sharp}A(B^{\pi}A)^{\sharp}+B^{\sharp}; \\ M_{4} =&-B^{\sharp}AB^{\sharp}. \end{aligned}$$

Proof

The proof is similar to Theorem 4, so we omit it. □

Next we consider a special case of Theorem 4 and 5, and investigate it over any ring.

Theorem 6

Let , where A,B∈R ^n×n. If B ^♯ exists and BAB ^π=0,B ^π AB=0, then

• (i) M ^♯ exists if and only if (AB ^π)^♯ exists.

• (ii) If M ^♯ exists, then

  $$M^{\sharp}=\left ( \begin{array}{c@{\quad}c}(AB^{\pi})^{\sharp}&B^{\sharp}\\ B^{\sharp}&-B^{\sharp }AB^{\sharp} \end{array} \right ). $$

Proof

The “only if” part of (i).

Let

$$M^{2}=\left ( \begin{array}{c@{\quad}c}I&AB^{\sharp}\\ 0&I \end{array} \right ) \left ( \begin{array}{c@{\quad}c}AB^{\pi}A+B^{2}&0\\ 0&B^{2} \end{array} \right ) \left ( \begin{array}{c@{\quad}c}I&O\\ B^{\sharp}A&I \end{array} \right ). $$

The decomposition of M is the same as in Theorem 4. By Lemma 2, M ^♯ exists if and only if there exist X,Y over R such that MX ²=M and YM ²=M.

Let

$$X=\left ( \begin{array}{c@{\quad}c}I&0\\ -B^{\sharp}A&I \end{array} \right ) \left ( \begin{array}{c@{\quad}c}X_{1}&X_{2}\\ X_{3}&X_{4} \end{array} \right ),\qquad Y=\left ( \begin{array}{c@{\quad}c}Y_{1}&Y_{2}\\ Y_{3}&Y_{4} \end{array} \right )\left ( \begin{array}{c@{\quad}c}I&-AB^{\sharp}\\ 0&I \end{array} \right ), $$

then we have

$$ AB^{\pi}=\bigl(AB^{\pi}A+B^2 \bigr)X_1, $$

(18)

$$ B=\bigl(AB^{\pi}A+B^{2}\bigr)X_2, $$

(19)

$$ B=B^{2}X_{3}, $$

(20)

$$ 0=B^{2}X_{4}, $$

(21)

$$ Y_{1}AB^{\pi}A+Y_{1}B^{2}=B^{\pi}A, $$

(22)

$$ Y_{2}B^{2}=B, $$

(23)

$$ Y_{3}AB^{\pi}A+Y_{3}B^{2}=B, $$

(24)

$$ Y_{4}B^{2}=0. $$

(25)

By (18) and (22), we have AB ^π AB ^π X ₁=B ^π AB ^π AB ^π X ₁=B ^π AB ^π AX ₁=B ^π AB ^π=AB ^π and Y ₁ AB ^π AB ^π=AB ^π, respectively. From Lemma 2, we know (AB ^π)^♯ exists. Next, we prove the sufficiency of (i) and the expression of (ii).

Let

$$X=\left ( \begin{array}{c@{\quad}c}(AB^{\pi})^{\sharp}&B^{\sharp}\\ B^{\sharp}&-B^{\sharp }AB^{\sharp} \end{array} \right ). $$

By Lemma 3,

$$MX=\left ( \begin{array}{c@{\quad}c}A(AB^{\pi})^{\sharp}+BB^{\sharp}&0\\ 0&BB^{\sharp} \end{array} \right )=XM. $$

It is easy to verify that XMX=M, MXM=M.

So X=M ^♯. □

Remark 2

We have expressed Theorems 1–2 and 4–5 over right Ore domains, but how to solve them in rings? This is still an open question.